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Not long ago someone sent me an email about my SubsetSum problem post, asking how to reconstruct the set of numbers that were used to create the sum. This is pretty straightforward, until you consider scenarios in which there is more than one correct subset and you want a specific one or all of them.

But, first thing’s first. Let’s consider the easy case – we assume there is only one correct result. For example consider the follwing set { 15, 20, 35, 40 } and the sum of 90 (15+35+40). Using the standard version of the algorithm, we’ll receive the table as follows (top row has values instead of indexes to increase readability, also, I skip non-relevant rows:

X

0

15

20

35

40

0

true

true

true

true

true

15

false

true

true

true

true

20

false

false

true

true

true

35

false

false

true

true

true

40

false

false

false

false

true

50

false

false

false

true

true

55

false

false

false

false

true

90

false

false

false

false

true

And so, to reconstruct values used, we’re going to determine which number was used to create each subset.

we start with our final value of 90, the first column in which we have ‘true’ is the last one, representing the value ’40’. We add 40 to the result set and substract this value from our final sum.And so recursively, we search for numbers until we reach 0. Starting from 90 we’ll find 40, then search for 90-40 = 50, 50 is true for column representing 35, so we’ll look for 50-35 = 15, we’ll find 15 in the second column and will not search for 15-15 = 0 (and we’re done)

The code can look something like this:

Determing the numbers used to create sum

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boolean[][]table=newboolean[sum+1][numbers.length+1];

Arrays.fill(table[0],true);

for(inti=1;i<=sum;i++){

for(intj=1;j<=numbers.length;j++){

table[i][j]=table[i][j-1];

if(numbers[j-1]<=i){

table[i][j]|=table[i-numbers[j-1]][j-1];

}

}

}

intcurrentSum=sum;

ArrayList<Integer>l=newArrayList<>();

while(currentSum>0){

for(intj=0;j<table[currentSum].length;j++){

if(table[currentSum][j]){

l.add(numbers[j-1]);

currentSum-=numbers[j-1];

break;

}

}

}

So what will happen if we want to find out what numbers to use to create 35? It can be either 15+20, or just 35. Algorithm in current form will return answer using as little numbers as possible , so it will be just 35. What to do to make it return both ?

Well, we need to remember in how many ways we can create any number from previous numbers. To do this, we’ll use int[][] instead of boolean[][] and increment the value each time we can find a new way to create a number

So the new table will look like this

X

0

15

20

35

40

0

1

1

1

1

1

15

0

1

1

1

1

20

0

0

1

1

1

35

0

0

1

2

2

We can see from this table that 35 can be created from either 35 (leading to 35-35 = 0 and end of algorithm) or from 20 (leading to 35-20 = 15).

We’ll need to branch on every such occasion, so it’s easiest to use recursion. It may look something like this:

getting all possible subsets

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publicstaticvoidprintAll(int[]numbers,int[][]t,intsum,

List<Integer>l){

if(sum==0){

System.out.println(l);

}else{

intcurrent=0;

for(inti=0;i<t[sum].length;i++){

if(t[sum][i]!=current){

current=t[sum][i];

intvalue=numbers[i-1];

l.add(value);

printAll(numbers,t,sum-value,l);

l.remove(l.size()-1);

}

}

}

}

publicstaticvoidgenerateSum(int[]numbers,intsum){

int[][]table=newint[sum+1][numbers.length+1];

Arrays.fill(table[0],1);

for(inti=1;i<=sum;i++){

for(intj=1;j<=numbers.length;j++){

table[i][j]=table[i][j-1];

if(numbers[j-1]<=i

&&table[i-numbers[j-1]][j-1]!=0){

table[i][j]++;

}

}

}

printAll(numbers,table,sum,newArrayList<Integer>());

}

There are two things I believe are worth noticing:

1. It’s relatively easy to apply the same optimizations as in the original post – directly if we have unique sums or more indirectly by creating lists of starting positions if we have multiple subsets giving the same sum.

2.It’s also relatively easy to count all possible subsets without enumerating them (just one line change, instead of table[i][j]++ do table[i][j] += table[i – numbers[j – 1]][j – 1];, table[sum].length – 1] will hold result after algorithm is done)

There is a lot of fun in modifying this to achieve results like ‘subset using least numbers’ or ‘subset minimizing the use of even numbers’, etc. most of which require just small changes. I encourage you to to play with it as a fun exercise.

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Recently I’ve encoundered a problem that required me to examine all possible partitions of a given set. Set partition is a collection of disjoint subsets whose union is the whole set. For example a set of {1,2,3} (being a partition by itself) can be devided in the following ways:

{1,2,3}

{1,2}, {3}

{1,3}, {2}

{1}, {2,3}

{1}, {2}, {3}

The problem was – how to generate those partitions efficiently. Moreover, we want to generate only the partitions where there is a specific number of sets. There is only one division where 3 or 1 sets are produced, and 3 partitions where we get 2 sets. The total number of possibilities for a set of given size is defined by Bell number

There is a connected problem to this one called Restricted growth functions (RGS) – such function is defined through inequality a[i] <= 1 + max(a[1]…a[i-1]) for i = 2,…,n. The reason those are connected is that you can express one by the other. A string produced by such function – for example 010 can be mapped to a set partition of {1,3}, {2} – first and third digits are zeroes, so first and third element of the set belong to subset zero, second digit is one, so the second set element belongs to the subset one, etc.

A string of 0121123 would express a set of {1},{2,4,5},{3,6},{7}.

The naive solution greatly repeats itself by not taking into account the mapping between Set partition and RGS.
Generating RGS is fairly easy:

Geenrating RGS

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publicstaticvoidgen(Strings,intlargest,intcurrentLength,

intdesiredLength){

if(currentLength==desiredLength){

System.out.println(s);

return;

}

for(inti=0;i<=largest;i++){

gen(s+i,i+1,currentLength+1,desiredLength);

}

}

Lazy generation and an iterative approach is somewhat more difficult – it relies on keeping track of a largest so far used value and iterating through all possibilities that satisfy the RGS inequality just as an Integer Partitioning algorithm would. I’ll refer you to the attached file for the implementation, just let me mention one important thing here that I find interesting.

This simple recursive approach generates all possible combinations that include any number of subsets from N 1-element to 1 N-element. Which is not exactly what I wanted.

I started out with an implementation that generated all the RGSes in an iterative manner, that also generated those permutations that didn’t satisfy the minGroups/maxGroup criteria (it skipped them, not returning from iterator). This worked fine, though was doing an awful lot of unnecessary skips. Generating 3 element subsets from 14 element set was taking 1,5 s.

First observation I made was that we can quickly determine whether we have enough groups in our permutation by examining the highest group used and if at any point there is exactly the number of ‘digits’ to generate as the desired number of groups minus currently highest used, then the ending digits are fixed (starting from current highest +1 in increasing order). This way we can ensure that we’ll always have the required number of groups.

The second observation regards the other end of the spectrum – maxGroup – all we need to do is to control whether the group number we’re assigning is not higher as the max number of groups and if so, disallow it just as we disallow any digits that would break the RGS inequality.

With those two observations acted upon in implementation we can generate only the RGSes that satisfy the conditions in linear time, thus reducing the example of 3 element subsets from 14 element set to 0,078s.

I attach the code for partition generating class without detailed explanations (other than ones above), if you find something difficult to understand (I’ve tried to keep the implementation concise), do let me know in comments and I’ll be sure to follow up.

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The Algorithm (the problem)

Yesterday I’ve encountered the following problem – from a set of positive integer numbers, determine all positive integers that can be constructed from any subset of that set.

For example set {1,3,5} can produce the following numbers {1,3,5,4,6,8,9}. In general case it’s a NP complete problem solvable by determining all combinations (this is what I implemented). However for anything larger than ~25 numbers in the set it’s strating to get too long to find the answer. Moreover, a ridiculous set like this {1,1,1,1,1,1,1,1,1,1,1,1…} that can be enumerated by fifth-grader becomes imposible to process.

So of course I went on looking for a smarter solution and quickly enough found this: http://en.wikipedia.org/wiki/Subset_sum_problem

Admittedly, yesterday was a long day, but I was sitting in front of the screen staring at this and couldn’t for the life of me understand how and why it works. So – let me put it out here is somewhat more elaborate but kind of step-by-step way (Once rested I understood it today and even managed to improve it a bit).

This isn’t exactly the problem as stated (find if subset sums to X or find all things subsets sum up to), but the algorithm used is the same.

How it works (on example)

Wikipedia states it’s pseudo-polynomial, the reason for it is simply that it doesn’t grow polynomialy with the number of items in the set, but with the range of numbers. So your problem with set like {1,2,3,4} will not take the same amount of time as set {1,2,3,99999999} (we’ll see why)

So – the general idea is to go through all numbers from 1 to sum({set}) (so for set {1,2,3,4} we’ll go through all numbers 1..(1+2+3+4 = 10)) and try to determine if this particular number can be constructed using first 1…X elements of the set. While trying to determine this we’ll construct an array storing the values for each number.

The array to store the results will be of type boolean and sizes [SUM_OF_NUMBERS_IN_SET+1][COUNT_OF_NUMBERS_IN_SET+1]

let’s work on an example – a trivial array of {1,3} the sum of elements is 4, the number of elements is 2.

We’ll produce an empty boolean array that will look like this

X

0

1

2

0

true

true

true

1

false

false

false

2

false

false

false

3

false

false

false

4

false

false

false

this is just initially set up table, the true,true in the first row indicates, that with no elements, or with only first item or with all items from the set, we can create a subset that sums up to zero (empty set) .

Now the algorithm kicks in – for each number (1…4) we’ll determine if with first 1…2 elements of the set, we can create this number. So – how can we determine that – there are just two simple cases – either we can create this number from a smaller subset (without including current number) or we can create this number from currently analysed number and we know that the remainder we can create from previously seen numbers. The boolean array will help us keep track of that. For each element, we’ll check if either an element to the left is true (first case) or an element higher up (by the amount of currently analysed element) and adjacent to left is true (second case).

In pseudocode that will be: table[i][j] = table[i][j-1] | table[i-set[j-1]][j-1] where set[j-1] is the value of element from the set (indexed from zero, hence j-1)

So – how will our example unfold:

row 1:

X

0

1

2

0

true

true

true

1

false

true

true

2

false

false

false

3

false

false

false

4

false

false

false

[1][1] we set to true because table[1-1][0] (that is table[0][0]) is true (second case)
[1][2] we set to true because table[1][1] is true (first case)

for next iteration, on row indexed by 2, we’ll not set anything to true:

[2][1] we’ll not set to true because neither [2][0] nor [1][0] is true
[2][2] we’ll not set to true because neither [2][1] nor [-1][0] is true

for the third row:

[3][1] we’ll not set to true because neither [3][0] nor [2][0] is true
[3][2] we’ll set to true because [0][1] is true

the last one is interesting :

[4][1] we’ll not set to true, because neither [4][0] nor [3][0] (this is the equivalent of saying, with just the first element from the set, we can’t construct 4)
[4][2] we’ll set to true because [1][1] is set to true (this is the equivalent of saying, with the first two elements, we can construct 4 – take 3, and somehow constructed 1 from first element from the set)

So the final table will look like this:

X

0

1

2

0

true

true

true

1

false

true

true

2

false

false

false

3

false

true

true

4

false

false

true

The answer to question ‘can we construct X’ from the set will be in table[X][set_size -1] (last element of the row)

Code

The actual code looks like this:

All subset sums

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intsum=0;

for(inti:numbers){

sum+=i;

}

boolean[][]table=newboolean[sum+1][numbers.length+1];

Arrays.fill(table[0],true);

for(inti=1;i<=sum;i++){

for(intj=1;j<=numbers.length;j++){

table[i][j]=table[i][j-1];

if(numbers[j-1]<=i){

table[i][j]|=table[i-numbers[j-1]][j-1];

}

}

}

And no we can print what values are possible to create from such set:

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for(intw=1;w<=sum;w++){

System.out.println(table[w][numbers.length]);

}

Improvement

The improvement isn’t that big (not big-O kind of thing), but in practical applications it matters. The main observation is that we only set the value to true at certain index for a number and then we keep it true. So we only need to track at which index we found we can make this number (if). The improved code (memory footprint is way smaller) can go like this:

Improved version of subset sum

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intsum=0;

for(inti:numbers){

sum+=i;

}

int[]starts=newint[sum+1];

Arrays.fill(starts,numbers.length+1);

starts[0]=0;

for(inti=1;i<=sum+1;i++){

for(intj=1;j<=numbers.length;j++){

intk=i-numbers[j-1];

if(k>=0&&starts[k]<j){

starts[i]=j;

break;

}

}

}

That’s it! Hope you’ll find this instructive and won’t waste too much time trying to understand how exactly are those indices set and why on a sunday evening.

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“Although the basic idea of binary search is comparatively straightforward, the details can be surprisingly tricky”. This statement is famously attributed to Donald Knuth, and I’ve recently learned the hard way how right he was.

Consider the following problem:

You’re given an inequality |A-X| < P – where A and P are some integer constants. Find how many integers satisfy that inequality. For simplicity let’s limit ourselves to X > – 2^31 and X < 2^31-1 Simple enough, we can divide the function |A-X| to two sections (- 2^31,A> where the function is strictly decreasing and <A,2^31-1) where it’s strictly increasing.

This can be solved easily analytically – X has to be between P-A and P+A, but let’s think about a generic monotonic inequality that can be specified on runtime.

To solve this more general problem,we only need to do a binary search for the minimum X satisfying inequality in the left section and maximum X satisfying inequality in the right one.

With the typical ‘wikipedia’ implementation of binary search by default we won’t do either (it assumes decreasing order, so it won’t do for the second section, and it returns invalid value if there isn’t an exact edge-value satisfying the inequality).

So there are two major considerations for our search function:

increasing / decreasing order of our search domain

edge values – should the function return the last value that satisfies conditions, the first one that doesn’t or some value indicating error.

For our example we need both increasing / decreasing order of search and last value that satisfies the conditions. Please note that in this case, we’re not really looking for a particular value (we don’t know what that value is). So, how can we achieve it? For example like this:

Binary Search

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publicstaticintsearch(intmin,intmax,

ISatisfactionValidator<Integer>validator,BinarySearchOrder order){

longl=min;

longr=max;

intrModif=0;

intlModif=0;

// choose which edge should be modified depending on the monotonicity

if(order==BinarySearchOrder.Increasing){

rModif=-1;

}else{

lModif=1;

}

do{

intm;

if(order==BinarySearchOrder.Increasing){

// need to make sure that in case of increasing sequences the

// right edge is selected if middle is between two values

m=(int)(l+(r-l+1)/2);

}else{

m=(int)(l+(r-l)/2);

}

// reverse the condition for icreasing - just to change the

// selection of sections for next pass

booleanvalid=validator.isValid(m)

^(order==BinarySearchOrder.Increasing);

if(valid){

r=m+rModif;

}else{

l=m+lModif;

}

}while(r>l);

return(int)l;

}

First a couple of explanations:

BinarySearchOrder is just an enum of Increasing/Decreasing (and is not that necessary as we’ll see later on). The ISatisfactionValidator is an interface with a single method isValid which takes an index (an X if you will) and returns true if the value produced for that X (whatever that would be) is still satisfies the conditions (whatever they may be) or not. This can easily be adapted to a single element search as we’ll see later on.

A couple of interesting points here:

long conversion for the calculation of middle point – this can be easily overlooked with disastrous consequences. M = L + (R-L)/2 – consider R = MAX and L = MIN (negative) – the R-L is way beyond 32-bits.

middle point selection depending on ordering – in case of increasing sequences we’re modifying the right edge so we need to make sure that whenever middle point is between two values, we choose the right one (ceil instead of floor)

finally – XOR for validity check is just to mirror the search pattern in case of increasing sequence.

One more note here – in case of multiple elements with the same value, in the decreasing sequence the algorithm will produce the first index (from the left) and in the increasing the last (first from right) with given value.

That is all well and good, but in this form, the algorithm deals nicely with monotonic functions and we usually deal with arrays. We can easily modify it to suit our needs:

Nothing really tricky here, we just determine the order and write our Stisfaction validator in such a way that it would be usable by the algorithm. The decision to throw an exception when an element is not found may in some cases not be the best, espectially in some performance critical code, but it can easy be modified.

Finally – just noticed right now that in generic search with all elements the same, a Decreasing option will be chosen (and therefore first element returned). It would be more appropriate to name the enum NonIncreasing (or modify the selection and choose Decreasing/NonDecreasing set)

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The other day I was watching a new series of lectures about Least significant digit Radix sort given by Robert Sedgewick and Kevin Wayne on Coursera (https://www.coursera.org/course/algs4partII) and it got me thinking – why don’t I use it anywhere ?

I got so comfy with Java’s mergesort and Comparable<T> that I never thought on optimizing that even in some performance critical code. So i though i should at least check what the performance impact can be.

LSD Radix sort is only useful in cases where you have a reasonable length of the keys on which you are sorting, so for my tests I assumed long as THE key data type. Reasoning behind was the following datatype:

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privatestaticclassMyData{

privateintid;

privateDate timestamp;

privateObjectblob;

...getters andsetters

}

since timestamp (Date object) can be expressed as long, and it’s perfectly reasonable to sort by date, it will do.

There are two additional things I had to define to enable more general sorting:

An interface that allows for extracting the sort key from object (and thus allowing for sorting various object by various keys – an equivalent of Comparator) and implementation of that for my data type.

We start by extracting the keys to a separate array (to avoid extracting them for every pass). It’s an optimization step, but it’s worth to do it, since calling the extractor method that many times will be expensive (but of course, there is a memory trade-off). Then, for each double octet (16 bits) starting from least significant ones, we do a key-index counting sort on extracted keys also moving the objects themselves into an auxiliary array. The result of each pass is an array sorted by that double byte. Because the sort is stable, we can start from least significant and move ‘up’.

The method is nor overly complicated, there are a couple of subtleties though:

lines 20-21 and 34-35 – the XOR with most significant bit of the most significant byte is required in order to correctly sort the negative numbers (negating the sign bit)

lines 40-46 – exchanging the auxiliary array with initial array (or just two aux arrays in case of keys) is needed in order to avoid copying all elements from aux array which is typically done in one-pass key-index sort.

Also note here, that thanks to the fact we have an even number of passes (4), we use the input array as an auxiliary array in the last pass, so we end up with sorted elements in original array and there is no need to copy elements from aux array to the original array.

So, is this any good ? Here’s results:

Size

Comparable

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10

0 ms.

10 ms.

100

1 ms.

9 ms.

1_000

3 ms.

6 ms.

10_000

37 ms.

10 ms.

100_000

178 ms.

65 ms.

1_000_000

466 ms.

179 ms.

10_000_000

8882 ms.

2219 ms.

For small number of elements it’s definitely not worth it, but that was expected. Above certain threshold though it definitely makes sense – it’s faster than system sort, and since Java uses mergesort for objects anyway, it doesn’t have that much bigger memory requirements than system sorting (the two aux key arrays, but in performance critical code if you want to sacrifice some purity for performance it too can be avoided).

There is one low-hanging optimization that can double the performance for large arrays – if you can limit your key size to a smaller number of bytes, you can get away with less passes. For example I tested with the Date converted to milliseconds from start of unix epoch which is long datatype, but afterwords did some tests with the key calculated by making the number of milliseconds relative to January 1-st 2000 (since I knew my data set couldn’t contain dates prior to that one) and lowered resolution to seconds which allowed me to fit it in 4 byte int.

It not always can be done, but as with many optimizations – a lot depends on inherent properties of your data.